use the references to access important values if needed for this question.\na 33.4 ml sample of a 0.518 m…

Explanation:

Step1: Calculate the moles of nitrous acid ($\ce{HNO2}$)

Use the formula $n = C\times V$ (where $n$ is moles, $C$ is concentration, $V$ is volume in liters). For $\ce{HNO2}$: $C = 0.518\ M$, $V=33.4\ mL=0.0334\ L$ $n_{\ce{HNO2}}=0.518\times0.0334 = 0.0173\ mol$

Step2: Calculate the moles of $\ce{Ba(OH)2}$

For $\ce{Ba(OH)2}$: $C = 0.372\ M$, $V = 14.7\ mL=0.0147\ L$ $n_{\ce{Ba(OH)2}}=0.372\times0.0147=0.00547\ mol$

Step3: Determine the moles of $\ce{OH-}$ from $\ce{Ba(OH)2}$

Since $1\ mol\ \ce{Ba(OH)2}$ gives $2\ mol\ \ce{OH-}$, $n_{\ce{OH-}}=2\times0.00547 = 0.0109\ mol$

Step4: Calculate the moles of $\ce{HNO2}$ remaining after reaction

The reaction is $\ce{HNO2 + OH- -> NO2- + H2O}$ Moles of $\ce{HNO2}$ remaining: $n_{\ce{HNO2 (remaining)}}=0.0173 - 0.0109=0.0064\ mol$

Step5: Calculate the total volume of the solution

$V_{total}=33.4 + 14.7=48.1\ mL = 0.0481\ L$

Step6: Calculate the concentration of $\ce{HNO2}$ remaining

$C_{\ce{HNO2}}=\frac{0.0064}{0.0481}\approx0.133\ M$

Step7: Use the Ka of $\ce{HNO2}$ ($Ka = 4.5\times 10^{-4}$) to find $[H+]$

For $\ce{HNO2 <=> H+ + NO2-}$, $Ka=\frac{[H+][NO2-]}{[HNO2]}$ Assume $[H+]=x$, then $[NO2-]=x$ and $[HNO2]=0.133 - x\approx0.133$ (since $x$ is small) $4.5\times 10^{-4}=\frac{x\times x}{0.133}$ $x^{2}=4.5\times 10^{-4}\times0.133$ $x^{2}=5.985\times 10^{-5}$ $x=\sqrt{5.985\times 10^{-5}}\approx7.74\times 10^{-3}\ M$

Step8: Calculate pH

$pH=-\log[H+]=-\log(7.74\times 10^{-3})\approx2.11$

Answer:

$2.11$